jesssica Group Title statics problem (: 2 years ago 2 years ago

1. jesssica Group Title

2. jesssica Group Title

im confused as to what forces are acting on points a and c.

3. Ishaan94 Group Title

Take components

4. jesssica Group Title

okay, so i did the sum of F in the y direction.. = Ay + Cy - W - T = 0 and then the sum of F in the x direction... = Ax + Cx - Fb = 0 then i did the moment about A and got Fb = 8.9, but i am stuck there.

5. jesssica Group Title

oh and Fb is acting in the negative x direction on point B.

6. Ishaan94 Group Title

At point A, it's the Normal reaction from Ground and the Tension from the string... at C it's just the string, is Pulley Mas-less and friction-less?

7. jesssica Group Title

i assumeee the pulley has no mass nor friction.

8. jesssica Group Title

wait so how many unknowns do i have ?

9. Ishaan94 Group Title

This is getting confusing for me :(

10. jesssica Group Title

me too! haha. and yay '94 kids !

11. jesssica Group Title

i am not sure what direction the force on point a is acting. perpendicular to the rod or the cable?

12. S Group Title

Does the answer 12.78 lb sound reasonable? I can't remember how do you do this, so I just did a bunch of stuff that seemed to be reasonable and I got that =\

13. jesssica Group Title

it does sound reasonable! but the back of the book has the answers, and evidently T = 9.08 lb ):

14. S Group Title

Hmmm.. Well at first i did what you did, sum of the forces in the y direction: Ay + Cy - T - 15 = 0

15. S Group Title

A is experiencing a force T, that's pulling on it at 10 degrees, so the y component of that force T that is acting on a would be sin(10) that would be equal to 0.1736T. So now 0.1736T + Cy - 1T - 15 = 0 -0.8264T + Cy - 15 = 0

16. S Group Title

Now if we can substitute something like that for Cy as well we could move on, but idk what to substitute for it cause there are no degrees for point C

17. jesssica Group Title

maybe we are overlooking something. ummmmm.

18. Ishaan94 Group Title
19. jesssica Group Title

20. Ishaan94 Group Title

$T \times \cos 10 \times 5 \times \sin 40 = \frac{5}{2} \times \cos 40 \times 15$ T is the tension in the cable. I used concept of Torques, one Torque is in effect due to its weight (rod's weight) the other is from the horizontal component of Cable's Tension (T * cos 10). Note. I calculated Toque about A.

21. S Group Title

Sum of x direction Ax - Fb = 0 Fb you said = 8.9 Ax - 8.9 = 0 Ax = 8.9 cos(10)T = 8.9 0.98T = 8.9 T = 9.08

22. jesssica Group Title

i havent learned torques, but s is something it looks like i can understand. thank you both !!

23. S Group Title

That kinda looks reasonable when you look at it. I think theres no Cx because its just like a continuation of Ax or something.. idk.. i can't remember.. I took it like a year and a half ago so i can't remember.. i'm gonna be a crappy engineer lol.. can't remember much lol

24. S Group Title

But at least it give the right answer.. i hope the procedure is right as well =)

25. jesssica Group Title

sike i still dont know what engineering even is and i am almost done with my first year of college. gracias yall !

26. S Group Title

Don't worry, you'll figure it out =D De nada =D